Jupiter2U School
Chapter 06

Measurement & Units

Middle School
At a glance
Core ideaEvery physical quantity is a number times a unit built from seven SI base units.
Key termDimensional analysis — checking an equation by its [M][L][T] dimensions.
You can…Test whether a formula is even possible before trusting it.
Watch outDimensions fix the form of a law but never the pure numbers (2π, ½).
Theory

Quantities, dimensions and the SI

Physics is quantitative: every statement about nature reduces to numbers attached to units. A physical quantity is written as a numerical value multiplied by a unit, e.g. v = 3.0 m·s⁻¹. The International System of Units (SI) defines seven base quantities from which all others are derived.

Base quantityUnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
Electric currentampereA
TemperaturekelvinK
Amount of substancemolemol
Luminous intensitycandelacd

Since 2019 the SI is defined by fixing exact numerical values of seven defining constants (e.g. the speed of light c = 299 792 458 m·s⁻¹, the Planck constant h, and the elementary charge e). Every other unit is derived by multiplying and dividing base units: force in kg·m·s⁻² (the newton, N), energy in kg·m²·s⁻² (the joule, J).

299 792 458
m·s⁻¹ — speed of light c (exact)
9.81
m·s⁻² — gravity g at Earth's surface
1.602×10⁻¹⁹
C — elementary charge e
6.674×10⁻¹¹
N·m²·kg⁻² — gravitation G
6.626×10⁻³⁴
J·s — Planck constant h
1.381×10⁻²³
J·K⁻¹ — Boltzmann kB

Dimensional analysis

Each quantity has a dimension — its expression in terms of mass [M], length [L] and time [T] (plus [I], [Θ], etc.). Speed has dimension [L][T]⁻¹; acceleration [L][T]⁻²; force [M][L][T]⁻². A valid equation must be dimensionally homogeneous: both sides share the same dimensions.

Explanation

Why units and significant figures matter

A number without a unit is meaningless in physics — "5" could be metres, seconds or apples. Units also carry information: because force is kg·m·s⁻², you can check any force formula by verifying its units come out in newtons. If they don't, the formula is wrong. Dimensional analysis is the cheapest error-detector in science.

Measurements are never exact. Significant figures communicate precision: 2.50 m claims certainty to the centimetre; 2.5 m only to the decimetre. When multiplying, the result keeps as many sig figs as the least precise input. Uncertainty is written x = (2.50 ± 0.02) m. Errors combine: for products, relative uncertainties add (Δz/z = Δx/x + Δy/y); for sums, absolute uncertainties add.

Step 1ObserveNotice a pattern or ask a precise, testable question about nature.
Step 2HypothesisePropose a model that predicts a measurable number.
Step 3ExperimentMeasure carefully, with units and quantified uncertainty.
Step 4AnalyseCompare prediction and data; check dimensions and sig figs.
Step 5RefineKeep, revise or reject the model — then repeat the loop.
Key idea. Orders of magnitude let you sanity-check anything. The universe spans roughly 10⁻³⁵ m (Planck length) to 10²⁶ m (observable universe) — 61 orders of magnitude. Estimating to the nearest power of ten (a "Fermi estimate") is a genuine physics skill.
Practical

Worked example — deriving the pendulum period by dimensions

Suppose the period T of a simple pendulum depends on its length ℓ, the bob mass m, and gravity g. Find the form of the relationship using dimensions alone.

  1. Assume T = k · ℓa mb gc, with k a dimensionless constant.
  2. Write dimensions: [T] = [L]a [M]b ([L][T]⁻²)c.
  3. Collect powers: [M]: b = 0; [L]: a + c = 0; [T]: 1 = −2c.
  4. Solve: c = −½, a = +½, b = 0. Mass drops out entirely.
  5. Conclude T = k √(ℓ/g). Full theory (Chapter 12) gives k = 2π.
T = 2π √(ℓ / g) simple pendulum, small angle

Check numerically: a 1.00 m pendulum with g = 9.81 m·s⁻² gives T = 2π√(1/9.81) = 2.01 s — famously close to 2 s, why the "seconds pendulum" is near 1 m.

θ ℓ T = 2π√(ℓ/g)
A pendulum's period depends only on its length ℓ and gravity g, not on the mass of the bob.
Q&A
Why can't dimensional analysis find the constant k?

Dimensionless factors like 2π, ½ or coefficients carry no dimensions, so they leave no trace in the [M][L][T] bookkeeping. Dimensional analysis fixes the form (which variables and with what powers) but never pure numbers. Those require full derivation or experiment.

Express the joule and the watt in SI base units.

Energy = force × distance, so J = N·m = kg·m²·s⁻². Power = energy / time, so the watt W = J·s⁻¹ = kg·m²·s⁻³.

A length is measured as (12.0 ± 0.1) cm and a time as (3.00 ± 0.05) s. What is the speed and its uncertainty?

Speed v = 0.120 m / 3.00 s = 0.0400 m·s⁻¹. Relative uncertainties add: Δv/v = 0.1/12.0 + 0.05/3.00 = 0.0083 + 0.0167 = 0.025. So Δv = 0.025 × 0.0400 = 0.0010 m·s⁻¹, giving v = (0.0400 ± 0.0010) m·s⁻¹.

Is the equation v² = u² + 2as dimensionally consistent?

v² and u² are [L]²[T]⁻². The term 2as is (acceleration × distance) = [L][T]⁻² · [L] = [L]²[T]⁻². All terms match, so yes — it is homogeneous (and indeed correct; see Chapter 7).

Concept mind map

How the ideas connect

Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.

SI base unitsDimensionsSig figsUnit conversionDimensionalanalysisUncertaintyMeasurement & Units
Infographic

The key facts, visualised

7 SI base units
metre, kilogram, second, ampere, kelvin, mole, candela
m/s^2
Unit of acceleration in SI
Sig figs
Digits that carry real precision
[L T^-1]
Dimensions of speed: length over time
Solved examples

Worked problems, step by step

Follow each solution line by line, then try to reproduce it on paper before moving on.

Example 1Convert 72 km/h into m/s.

  1. 1 km = 1000 m and 1 h = 3600 s
  2. 72 km/h = 72 * 1000 / 3600 m/s
  3. = 72000 / 3600

Example 2Use dimensions to check if T = 2*pi*sqrt(L/g) is valid.

  1. L has dimension [L], g has [L T^-2]
  2. L/g has [T^2], sqrt gives [T]
  3. Both sides have dimension of time
Practice problem set

Now you try

Work each one out first, then tap to reveal the worked answer.

1How many significant figures are in 0.00420?
Three significant figures: 4, 2 and the trailing 0.
2What are the SI base units of force?
Force in newtons is kg*m/s^2, from mass and acceleration.
3Convert 2.5 hours into seconds.
2.5 * 3600 = 9000 s.
4What is the dimension of area?
Length squared, [L^2], with SI unit m^2.
5Round 3.14159 to three significant figures.
3.14.
6Why do we use dimensional analysis?
To check equations are consistent and to derive relationships between quantities.